Find the limit as $x$ approaches negative infinity. $\lim_{x\to-\infty}\dfrac{9x^6}{\sqrt{9x^{12}+4x^6}}=$
Explanation: Let's find this limit directly. To do that, we will want to divide both the numerator and the denominator by the same quantity, in a way that will help us derive the limit. Since the leading term of the numerator is $x^6$, let's divide by $x^6$. In the denominator, let's divide by $\sqrt{x^{12}}$, since for any value, $x^6=\sqrt{x^{12}}$. $\begin{aligned} &\phantom{=}\lim_{x\to-\infty}\dfrac{9x^6}{\sqrt{9x^{12}+4x^6}} \\\\ &=\lim_{x\to-\infty}\dfrac{\dfrac{9x^6}{x^6}}{\dfrac{\sqrt{9x^{12}+4x^6}}{\sqrt{x^{12}}}} \gray{\text{Divide sides by }x^6=\sqrt{x^{12}}} \end{aligned}$ Now let's continue by simplifying the expression and using the fact that for any nonzero number $k$ and positive power $n$, the limit $\lim_{x\to-\infty}\dfrac{k}{x^n}$ is equal to $0$. $\begin{aligned} &=\lim_{x\to-\infty}\dfrac{\dfrac{9\cancel{x^6}}{\cancel{x^6}}}{\sqrt{\dfrac{9\cancel{x^{12}}}{\cancel{x^{12}}}+\dfrac{4\cancel {x^6}}{\cancel {x^6}\cdot x^6}}} \\\\ &=\lim_{x\to-\infty}\dfrac{9}{\sqrt{9+\dfrac{4}{x^6}}} \\\\ &=\lim_{x\to-\infty}\dfrac{9}{\sqrt{9+0}} \gray{\lim_{x\to-\infty}\dfrac{k}{x^n}=0} \\\\ &=\dfrac{9}{\sqrt{9}} \\\\ &=\dfrac{9}{3} \\\\ &=3 \end{aligned}$ In conclusion, $\lim_{x\to-\infty}\dfrac{9x^6}{\sqrt{9x^{12}+4x^6}}=3$.